hdu 1024 Max Sum Plus Plus（动态规划+m子段和的最大值）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1024

Max Sum Plus Plus

Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22262    Accepted Submission(s): 7484

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S ₁, S ₂, S ₃, S ₄ ... S _x, ... S _n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S _x ≤ 32767). We define a function sum(i, j) = S _i + ... + S _j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + sum(i _m, j _m) maximal (i _x ≤ i _y ≤ j _x or i _x ≤ j _y ≤ j _x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i _x, j _x)(1 ≤ x ≤ m) instead. ^_^

 

Input

Each test case will begin with two integers m and n, followed by n integers S ₁, S ₂, S ₃ ... S _n.
Process to the end of file.

 

Output

Output the maximal summation described above in one line.

 

Sample Input

1 3 1 2 3 2 6 -1 4 -2 3 -2 3

 

Sample Output

6 8

Hint

Huge input, scanf and dynamic programming is recommended.

 

Author

JGShining（极光炫影）

 

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题目大意：输入一个m，n分别表示成m组，一共有n个数即将n个数分成m组，m组的和加起来得到最大值并输出。

解题思路：状态dp[i][j]表示前j个数分成i组的最大值。

动态转移方程：dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) (0

dp[i][j-1]+a[j]表示的是前j-1分成i组，第j个必须放在前一组里面。

max( dp[i-1][k] ) + a[j] )表示的前（0

但是题目的数据量比较到，时间复杂度为n^3，n<=1000000，显然会超时，继续优化。

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个
的最大值 用数组保存下来 ，这样时间复杂度为 n^2。

详见代码。

#include 
#include 
#include 

using namespace std;

#define N 1000000
#define INF 0x7fffffff

int a[N+10];
int dp[N+10],Max[N+10];//max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。

int main()
{
    int n,m,mmax;
    while (~scanf("%d%d",&m,&n))
    {
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        memset(Max,0,sizeof(Max));
        for (int i=1;i<=m;i++)//分成几组
        {
            mmax=-INF;
            for (int j=i;j<=n;j++)//j个数分成i组，至少要有i个数
            {
                dp[j]=max(dp[j-1]+a[j],Max[j-1]+a[j]);
                Max[j-1]=mmax;
                mmax=max(mmax,dp[j]);
            }
        }
        printf ("%d\n",mmax);
    }
    return 0;
}